Question

The difference between C.I and S.I on a certain sum of money invested for 3 years at 6% p.a
is 110.16. The sum is
(a) * 3,000 (b) * 3,700
(c) 12,000 (d) 10,000​

Answer 1

AnswEr :

$\bullet\:\textsf{Difference between CI and SI = Rs. 110.16}\\\bullet\:\textsf{Time = 3 years}\\\bullet\:\textsf{Rate = 6 \% p.a.}$

$\rule{150}{2}$

$\underline{\bigstar\:\textsf{Difference b/w CI and SI for 3 years :}}$

$:\implies\sf Difference = \dfrac{Pr^2(300+r)}{100^3}\\\\\\:\implies\sf110.16=\dfrac{P(6)^2(300+6)}{100^2 \times 100}\\\\\\:\implies\sf110.16 \times 100=\dfrac{P \times 36 \times 306}{100^2}\\\\\\:\implies\sf \dfrac{ \cancel{11016} \times 10000}{\cancel{36} \times 306}=P\\\\\\:\implies\sf P =\dfrac{\cancel{306} \times 10000}{\cancel{306}}\\\\\\:\implies \boxed{\sf Principal = Rs. \: 10000}$

⠀

$\underline{\therefore\:\textsf{The Required Sum will be ( d) \textbf{Rs. 10000}}}$

Answer 2

$\huge{\underline{\underline{\red{\mathfrak{AnSwEr :}}}}}$

$\small{\underline{\blue{\sf{Given :}}}}$

• Difference between CI and SI = Rs. 110.16
• Time = 3 years
• Rate = 6 %

$\rule{200}{1}$

$\small{\underline{\green{\sf{Solution :}}}}$

$\large \star {\boxed{\sf{Difference = \dfrac{Pr^2(300+r)}{100^3}}}} \\ \\ \implies {\sf{110.16 \: = \: \dfrac{P(6)^2 \big(300 \: + \: 6\big)}{100^3}}} \\ \\ \implies {\sf{110.16 \: = \: \dfrac{P(36) \big( 306 \big) }{100^3}}} \\ \\ \implies {\sf{110.16 \: = \: \dfrac{36P \: \times \: 306}{100^3}}} \\ \\ \implies {\sf{110.16 \: \times \: 100^3 \: = \: 11016 \: \times \: P}} \\ \\ \implies {\sf{P \: = \: \dfrac{\cancel{11061} \: \times \: 100^2}{\cancel{11061}}}} \\ \\ \implies {\sf{P \: = \: 100^2}} \\ \\ \implies {\boxed{\sf{P \: = \: Rs. 10000}}}$

∴ Answer is (d) Rs. 10000